On: Blue Eyes Logic Puzzle

[Mindless2112]

Hmm. That seems right. With 3 blue-eyed people, a blue-eyed person observing that the 2 blue-eyed people did nothing does add the additional knowledge that there are a total of 3 blue-eyed people.

So the additional knowledge added for 3 or more blue-eyed people is a lower bound on the total number of blue-eyed people (which grows over time), but it's not exactly clear to me why the foreigner's statement triggers this.

[vytasgd → Mindless2112]

I've been with you on this, but I think we are wrong. Here's where I'm at: (without foreigner's statement)

1 blue eyed person- He know's nothing, nothing happens

2 blue eyed people- Each thinks the other could be the only blue but they don't know it... deadlock.

3 blue eyed people- Each thinks the other two are caught in the 2 person deadlock scenario knowing nothing.

4 blue eyed people- Each thinks the other 3 are caught in the 3 person scenario etc...

Once the foreigner adds the knowledge:

1- he'd leave the first day

2- the second would recognize the first didn't leave, they'd both leave on day 2

3- the third would recognize the first two didn't leave on day 2, etc, etc, we were wrong...bollocks.

[pavelrub → Mindless2112]

The new information (common knowledge) is added immediately to everybody, not after 2 days. Using this information (that everybody knows that everybody knows that everybody knows...), it is then possible to decide whether you have blue eyes or not after 2 days.

[DannoHung → Mindless2112]

I dunno, but it's interesting that if you think about it, even if he said to one person, "You have blue eyes." Only that one person would do anything.

Or, the other interesting thing is what if there were 500 and 500?

It's a very weird logic puzzle for sure.